# Difficult contour integral.

1. May 22, 2015

### demoncore

• Missing homework template due to originally being posted in other forum.
I am attempting to calculate the following integral.

$$\frac{1}{2\pi i}\int_C \frac{du}{u^2} exp({-\frac{(q - \frac{q_0}{2i} (u - u^{-1}))^2}{2\sigma^2}})$$

Taken over the unit disk. I first make the substitution $$z = q - \frac{q_0}{2i} (u - u^{-1})$$ So,
$$dz = -\frac{q_0}{2i}(1 + u^{-2})du$$

When I attempt to back-substitute in for u, however, I find the following expression:

$$u = \frac{(q - z)i \pm \sqrt{q_0^2 - (q - z)^2}}{q_0}$$
I am not sure where to proceed from here. Arbitrarily choosing one or the other solution for u doesn't seem to give me the correct answer. Any help would be appreciated.

2. May 23, 2015

### ecastro

Have you tried deconstructing your $e$ to multiple terms? I think you could take a term outside the integral.

3. May 23, 2015

### Ray Vickson

If you set $u = e^{i \theta}$ you can write the integral (call it $J\,$ ) in the form
$$J = \frac{1}{2 \pi i} \int_0^{2 \pi} i e^{i \theta} e^{-2 i \theta} E(\theta) \, d \theta,\\ E(\theta) = \exp \left(- \frac{1}{2 \sigma^2} \left(q - q_0 \sin(\theta) \right)^2 \right)$$
Thus, $J = J_r+ i J_i$, where
$$J_r = \frac{1}{2 \pi} \int_0^{2 \pi} \cos(\theta) E(\theta) \, d \theta, \\ J_i = -\frac{1}{2 \pi} \int_0^{2 \pi} \sin(\theta) E(\theta) \, d \theta$$
Note that $J_r = 0$ because the integrand is asymmetric about $\theta = \pi/2$ on the interval $[0,\pi]$ and is asymmetric about $\theta = 3 \pi/2$ on the interval $[\pi, 2\pi]$. However, $J_i \neq 0$. In numerous numerical examples we find $J_i < 0$.

I doubt that $J_i$ has a simple (if any) elementary formula, but numerical integration works well on it, especially if one computes the parts for $[0,\pi]$ and $[\pi, 2\pi]$ separately. Alternatively, this separation can be done analytically, to get
$$J_i = \frac{1}{2 \pi} \int_0^{\pi} \sin(\theta) F(\theta), \, d \theta, \\ F(\theta) = \exp \left(- \frac{1}{2 \sigma^2} (q + q_0 \sin(\theta))^2 \right) - \exp \left(- \frac{1}{2 \sigma^2} (q - q_0 \sin(\theta))^2 \right)$$